In an isosceles triangle, the height drawn to the base is 16 and the radius of the inscribed circle is 6

In an isosceles triangle, the height drawn to the base is 16 and the radius of the inscribed circle is 6, find the perimeter of the triangle.

Let’s draw the radius of the circle OK, which is perpendicular to the side AB, then the triangle OBK is rectangular.

The length of the height AH = 16 cm, and OH = 6 cm as the radius of the circle, then OB = 16 – 6 = 10 cm.

From the triangle BOK BK ^ 2 = OВ ^ 2 – OK ^ 2 = 100 – 36 = 64.

BK = 8 cm.

AK = AH as tangents to the circle drawn from one point. Let AK = AH = X cm.

Then AB = X + 8 cm.

From a right-angled triangle ABH, AB ^ 2 = BH ^ 2 + AH ^ 2.

(X + 8) ^ 2 = 256 + X ^ 2.

X ^ 2 + 16 * X + 64 = 256 + 2.

16 * X = 192.

X = AH = AK = 192/16 = 12 cm.

Then AB = BC = 8 + 12 = 20 cm.

AC = 2 * AH = 2 * 12 = 24 cm.

Determine the perimeter of the triangle. Ravs = AB + BC + AC = 20 + 20 + 24 = 64 cm.

Answer: The perimeter of the triangle is 64 cm.

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